Ponder This

August 2013

To solve this month's challenge, you can try various types of searches, but a nice and elegant solution is to use the following linear matrix A:

100001101
010001011
001000111
000100001
000010001

and assign each nine-bit vertex v the five-bit value of the result of the matrix multiplication Av.
It is easy to show that the first (v=0) vertex has all 32 types of neighbors. Lets call them v_0, v_1,..., v_31.
From this follows that if we want to go from any vertex w of color c_1 to a neighbor of color c_2, all we need is to take the w xor v_{c_1 xor c_2} vertex.

Don Dodson suggested a nice solution that not only contains the name and date but also the word "IBM" (look down and you'll find it vertically).